LeetCode2-两数相加

大佬解法

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) { val = x; }* }*/
class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode pre = new ListNode(0);ListNode cur = pre;int carry = 0;while(l1 != null || l2 != null) {int x = l1 == null ? 0 : l1.val;int y = l2 == null ? 0 : l2.val;int sum = x + y + carry;carry = sum / 10;sum = sum % 10;cur.next = new ListNode(sum);cur = cur.next;if(l1 != null)l1 = l1.next;if(l2 != null)l2 = l2.next;}if(carry == 1) {cur.next = new ListNode(carry);}return pre.next;}
}

自己的解法

写的跟屎一样,虽然击败100%Java用户,但是写的实在是太过于复杂

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {int sum=0;ListNode ans=new ListNode();ListNode p=ans,ptr=null;for(;;){if(l1!=null&&l2!=null){sum=l1.val+l2.val;if(p.val==1){sum+=1;}if(sum>=10){p.next=new ListNode(1);p.val=sum%10;}else{p.val=sum;if(l1.next!=null||l2.next!=null)//防止产生前导0{p.next=new ListNode();}}l1=l1.next;l2=l2.next;ptr=l1==null?l2:l1;p=p.next;}else if(ptr!=null){int null_sum=0;if(p.val==1){null_sum=ptr.val+p.val;if(null_sum>=10){p.next=new ListNode(1);p.val=null_sum%10;}else{p.val=null_sum;if(ptr.next!=null){//防止产生前导0p.next=new ListNode();}}}else{p.val=ptr.val;if(ptr.next!=null){//防止产生前导0p.next=new ListNode();}}//往后遍历ptr=ptr.next;p=p.next;}else{break;}}return ans; }}